Question

Let there are odd number of stones placed one by one at intervals of $10\text{m}$ along a road. These stones have to be assembled around the middle stone. A person can carry only one stone at a time. A man carried out the job with one of the end stones by carrying them in succession. In carrying all the stones, he covered a distance of $3\text{km}$. Then the number of stones is

• A. $12$
• B. $21$
• C. $25$
• D. $31$

Answer 1

The correct option is C $25$

Let there be $2n+1$ stones. So, there will be $n$ stones each side of middle stone.
Suppose man started job from middle stone.
The distance covered for first stone $=2\times 10\text{m}$
(as he has to go back to pick the stone and come back)

Distance covered for second stone $=2\times 2\times 10\text{m}$

So, distance covering for one side is
$=2\times 10+2\times 2\times 10+2\times 3\times 10+\cdots +2\times n\times 10$
$=2[10+2\times 10+3\times 10+\cdots +n\times 10]$

Same thing is repeated for the second side. So, same distance is cover for second side.

Since, man started job from one end,
Therefore, total distance covered is given by,
$=4[10+2\times 10+3\times 10+...+n\times 10]-n\times 10$
$=40[1+2+3+...+n]-10n$
$=20n^2+10n$

But, the total distance covered $=3\text{km}=3000\text{m}$

$\therefore 20n^2+10n=3000$
$\Rightarrow 2n^2+n-300=0$
$\Rightarrow (n-12)(2n+25)=0$
$\Rightarrow n=12$

Hence, number of stones is $2n+1=25$.